Real Analysis Folland, Proposition 6.13, The Dual of $L^p$ 
Proposition 6.13 - Suppose $p$ and $q$ are conjugate exponents and $1\leq q < \infty$. If $g\in L^q$, then 
  $$\|g\|_{q} = \|\phi_g\| = \sup\{\left|\int fg\right|: \|f\|_{p} = 1 \}$$
  If $\mu$ is semifinite, this result also holds for $q = \infty$.

Part of Proof - 
We know that for Holder's inequality that $\|\phi_g\| \leq \|g\|_{q}$, equality is trivial if $g = 0$ a.e. If $g\neq 0$ and $q < \infty$, let 
$$f = \frac{|g|^{q-1}\overline{sgn g}}{\|g\|_{q}^{q-1}}$$
Then,
$$\|f\|_{p} = \frac{\int |g|^{(q-1)p}}{\|g\|_{q}^{(q-1)p}} = \frac{\int |g|^q}{\int |g|^q} = 1$$
So,
$$\|\phi_g\| \geq \int fg $$
I don't understand why this is the case? Any insight or suggestions is greatly appreciated
 A: 
Proposition 6.13 - Suppose $p$ and $q$ are conjugate exponents and $1\leq q < \infty$. If $g\in L^q$, then 
  $$\|g\|_{q} = \|\phi_g\| = \sup\{\left|\int fg\right|: \|f\|_{p} = 1 \}$$
  If $\mu$ is semifinite, this result also holds for $q = \infty$.

Proof - 
Note that for $1\leq q \leq  \infty$, we know, for Holder's inequality, that $\|\phi_g\| \leq \|g\|_{q}$, because for all $f \in L^p$ such that $\|f\|_p=1$, we have
$$\left|\int fg\right| \leq  \|f\|_p\|g\|_{q}=\|g\|_{q}$$
So 
$$ \|\phi_g\| = \sup\left \{\left|\int fg\right|: \|f\|_{p} = 1 \right \} \leq \|g\|_{q} \tag{1} $$
Note that the equality is trivial if $\|g\|_{q} = 0$. Now, if $\|g\|_{q}\neq 0$ and $q < \infty$, let 
$$h = \frac{|g|^{q-1}\overline{\textrm{sgn} g}}{\|g\|_{q}^{q-1}}$$
Then, (and it is important) let us prove that such $h$ is in $L^p$ and $\|h\|_{p}=1$. In fact, we have
$$\|h\|_{p}^p =\int | h |^p=  \frac{\int |g|^{(q-1)p}}{\|g\|_{q}^{(q-1)p}} = \frac{\int |g|^q}{\int |g|^q} = 1$$
Note that 
$$\int hg =  \frac{\int |g|^{(q-1)}(\overline{\textrm{sgn} g})g}{\|g\|_{q}^{(q-1)}} = \frac{\int |g|^{(q-1)}| g|}{\|g\|_{q}^{(q-1)}}=  \frac{\int |g|^{q}}{\|g\|_{q}^{(q-1)}}= \frac{\|g\|_{q}^{q}}{\|g\|_{q}^{(q-1)}}=\|g\|_{q} $$
So,
$$ \|g\|_{q} = \int hg = \left |  \int hg  \right | \leq \sup\left \{\left|\int fg\right|: \|f\|_{p} = 1 \right \} = \|\phi_g\|$$
So, for $q<\infty$, using $(1)$ we have 
$$\|g\|_{q} = \|\phi_g\| = \sup \left \{\left|\int fg\right|: \|f\|_{p} = 1 \right\}$$
Now suppose $q=\infty$ and $\mu$ is semifinite. 
For any $\epsilon>0$, let $A =\{x: |g(x)| >\|g\|_\infty -\epsilon\}$. Then $\mu(A)>0$, so since $\mu$ is semifinite, there exists $B \subset A$ such that $0<\mu(B) <\infty$. Let 
$$h= \frac{\chi_B \overline{\textrm{sgn}g}}{\mu(B)}$$
Then 
$$\int |h| d\mu = \int \frac{\chi_B | \overline{\textrm{sgn}g} |}{\mu(B)}d\mu =  \int \frac{\chi_B }{\mu(B)}d\mu =1$$
So $h\in L^1$ and $\|h\|_1=1$. And we have 
$$ \int h g d\mu = \frac{ \int\chi_B ( \overline{\textrm{sgn}g})g d\mu}{\mu(B)}=  \frac{ \int\chi_B |g|d\mu}{\mu(B)}\geq \|g\|_\infty-\epsilon $$
So we have 
$$  \|g\|_\infty-\epsilon \leq \int h g d\mu \leq \left| \int h g d\mu   \right|  \leq \sup\left \{\left|\int fg\right|: \|f\|_{1} = 1 \right \} = \|\phi_g\|$$
So, using $(1)$ for $q=\infty$, we have 
$$\|g\|_{\infty} = \|\phi_g\| = \sup \left \{\left|\int fg\right|: \|f\|_{1} = 1 \right\}$$
A: In fact we always define $g^*$ as the conjugate function of $g \in L^q$ by 
\begin{equation*}
g^* = \Vert g \Vert_q^{1-q} \cdot \textrm{sgn}g \cdot |g|^{q-1}.
\end{equation*} 
It's easy to show that 
\begin{equation*}
\int g\cdot g^* = \Vert g \Vert_q \text{ and } \Vert g^* \Vert_p = 1,
\end{equation*}
 which shows that 
\begin{equation*}
\sup\left\{\left|\int fg\right|:f \in L^p, \Vert f \Vert_p = 1\right\} \geqslant \Vert g \Vert_q.
\end{equation*}
